What is 2e^{-(t-1)u(t-1)} ?

The solution to 2e^{-(t-1)u(t-1)} is 2/(e^{u(t-1)^2)}

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simplify 2e^{-(t-1)u(t-1)}

simplify

2 e^{- (t - 1) u (t - 1)}

\frac{2}{e ^{u (t - 1)^{2}}}

Solution steps

2 e^{- (t - 1) u (t - 1)}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= 2 \cdot \frac{1}{e ^{u (t - 1) (t - 1)}}

\frac{1}{e ^{(t - 1) u (t - 1)}} = \frac{1}{e ^{u (t - 1)^{2}}}

= 2 \cdot \frac{1}{e ^{u (t - 1)^{2}}}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{e ^{u (t - 1)^{2}}}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{e ^{u (t - 1)^{2}}}

3.7 x 1 0^{- 4}

y^{2 + a} e^{- l n (y^{a})}

(- 1)^{6} e^{- (- 1)^{2}}

\frac{- 90}{45}

What is 2e^{-(t-1)u(t-1)} ?
The solution to 2e^{-(t-1)u(t-1)} is 2/(e^{u(t-1)^2)}